Question

Cho P = \(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{x\sqrt{x}+\sqrt{x}-2x^2}{x+1}\) ( với \(x\) > 0 )

a. Rút gọn P

b. Tìm x sao cho P > 2

c.Tìm x thuộc Z sao cho P thuộc Z

Answer 1

a,\(ĐKXĐ:x\in R|x>0\)( Vì cả 2 mẫu đều >0)

Xét:\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}+1\right)=x+\sqrt{x}\)

\(\Rightarrow\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{x\sqrt{x}+\sqrt{x}-2x^2}{x+1}=\dfrac{3x^2+x}{x+1}=P\\ \)

b, \(P< 2\Leftrightarrow\dfrac{3x^2+x}{x+1}< 2\Rightarrow3x^2-x-2< 0\Rightarrow\left(3x+2\right)\left(x-1\right)< 0\Rightarrow-\dfrac{2}{3}< x< 1\\ \)c, \(P\in Z\Leftrightarrow\dfrac{3x^2+x}{x+1}\in Z\Leftrightarrow3x^2+x⋮x+1\)

Tự lm nốt nhé.